Question

In Problems 1–18 use Definition 7.1.1 to find$L\left\{f\left(t\right)\right\}.$

6.$f\left(t\right)={\{}_{\cos t}^{0,}\raisebox{1ex}{$\raisebox{1ex}{$0\le t\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$}\!\left/ \!\raisebox{-1ex}{$\raisebox{1ex}{$t\ge \pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$}\right.$

Short answer

The Laplace transform of above function is,

$I=\frac{-e^{-\frac{s.\pi }{2}}}{\left(1+s^2\right)}$

Step-by-step solution

Definition 7.1.1 Laplace transform

Let f be a function define for$t⩾0$.Then the integral

$L\left\{f\left(t\right)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$

is said to be Laplace transform of f provide that integral converges.

Applying the definition

Consider the function $f\left(t\right)=\left\{{}_{cost,t\ge \frac{\pi }{2}}^{0,0\le t\le \frac{\pi }{2}}\right.$

The objective is to find $L\left\{f\left(t\right)\right\}$using the definition.

Note that, the function f is defined for$t⩾0$.

From the definition,

$L\left\{f\left(t\right)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$

Sincerole="math" localid="1663940424049" $f$is defined in two pieces$[0,\frac{\pi }{2})$and $[\frac{\pi }{2},\infty )$Laplacian if f is$L\left\{f\left(t\right)\right\}$expressed as the sum of two integrals.

$L\left\{f\left(t\right)\right\}={\int }_0^{\frac{\pi }{2}}{e}^{-st}f\left(t\right)dt+{\int }_{\frac{\pi }{2}}^{\infty }{e}^{-st}f\left(t\right)dt$

$={\int }_0^{\frac{\pi }{2}}{e}^{-st}\left(0\right)dt+{\int }_{\frac{\pi }{2}}^{\infty }{e}^{-st}\left(cost\right)dt$

$I=0+{\int }_{\frac{\pi }{2}}^{\infty }\cos t.e^{-st}dt$

Let solve first, by using integral by parts formula

Soformula is$I=\int u.vdt=u\int vdt-\int \left[\frac{d}{dt}u.\int vdt\right]dt$

Where u and v we choose according to ILATE rule;

I= Inverse

L= Logarithmic

A= Arithmetic

T= Trigonometry

E= Exponential

Asarithmeticfunctioncomesfirst,

Therefore,

$$\begin{gathered} I={\int }_{\frac{\pi }{2}}^{\infty }\cos t.e^{-st}dt \\ u=e^{-st};v=cost \\ I=e^{-st}\int \cos tdt-\int \left[\frac{d}{dt}{e}^{-st}.\int \cos tdt\right]dt \\ ={\left\{e^{-st}.\sin t\right\}}_{\frac{\pi }{2}}^{\infty }-\int \left\{-s.e^{-st}.\sin t\right\}dt \\ ={\left\{e^{-st}.\sin t\right\}}_{\frac{{\displaystyle \pi }}{{\displaystyle 2}}}^{\infty }+s\int \left\{e^{-st}.\sin t\right\}dt \\ ={\left\{e^{-st}.\sin t\right\}}_{\frac{{\displaystyle \pi }}{{\displaystyle 2}}}^{\infty }+s.I_1 \end{gathered}$$

Againweuseintegralbypartsformulatosolveit,

$$\begin{gathered} I_1=\int e^{-st}\sin tdt \\ u=e^{-st};v=\sin t \\ I=e^{-st}\int \sin tdt-\int \left\{\frac{d}{dt}{e}^{-st}\int \sin tdt\right\}dt \\ ={\left\{-e^{-st}.\cos t\right\}}_{\frac{\pi }{2}}^{\infty }-{\int }_{\frac{\pi }{2}}^{\infty }s.e^{-st}.costdt \\ ={\left\{-e^{-st}.\cos t\right\}}_{\frac{\pi }{2}}^{\infty }-s{\int }_{\frac{\pi }{2}}^{\infty }{e}^{-st}.\cos tdt \\ {I}_1={\left\{-e^{-st}.\cos t\right\}}_{\frac{{\displaystyle \pi }}{{\displaystyle 2}}}^{\infty }-s.I \end{gathered}$$

Simplification

Wecombineand for further simplification to get required integral;

$$\begin{gathered} I={\left\{e^{-st}.\sin t\right\}}_{\frac{\pi }{2}}^{\infty }+\left\{{\left(-s.e^{-st}\cos t\right)}_{\frac{\pi }{2}}^{\infty }\right\}-s^{2}I \\ I+s^{2}I={\left\{e^{-st}.sint\right\}}_{\frac{\pi }{2}}^{\infty }+{\left\{-s.e^{-st}\cos t\right\}}_{\frac{\pi }{2}}^{\infty } \\ I=\frac{1}{\left(1+s^2\right)}{\left\{e^{-st}.\left(sint-s.\cos t\right)\right\}}_{\frac{\pi }{2}}^{\infty } \\ I=\frac{1}{\left(1+s^2\right)}\left\{\left[e^{-s.\infty }.\left(\sin \infty -s.\cos \pi \right)\right]-\left[e^{-s.\frac{\pi }{2}}.\left(\sin \frac{\pi }{2}-s.\cos \frac{\pi }{2}\right)\right]\right\} \\ I=\frac{1}{\left(1+s^2\right)}\left\{0-\left[e^{-\frac{s\pi }{2}}.\left(1-s.0\right)\right]\right\} \\ I=\frac{-e^{-\frac{s.\pi }{2}}}{\left(1+s^2\right)} \end{gathered}$$

Therefore the required Laplace transform of function is,

$I=\frac{-e^{-\frac{s.\pi }{2}}}{\left(1+s^2\right)}$